Question

Consider two hot bodies $B_1$ and $B_2$ which have temperatures ${100}^{\circ }\text{C}$ and ${80}^{\circ }\text{C}$, respectively at $t=0$. The temperature of the surroundings is ${40}^{\circ }\text{C}$. The ratio of the respective rates of cooling $R_1$ and $R_2$ of these two bodies at $t=0$ will be

• A. $R_1:R_2=3:2$
• B. $R_1:R_2=5:4$
• C. $R_1:R_2=2:3$
• D. $R_1:R_2=4:5$

Answer 1

The correct option is A $R_1:R_2=3:2$

Given that,
Temperature of body $B_1,\left({\theta }_1\right)={100}^{\circ }\text{C}$
Temperature of body $B_2,\left({\theta }_2\right)={80}^{\circ }\text{C}$
Temperature of the surroundings $\left({\theta }_0\right)={40}^{\circ }\text{C}$

According to Newton's law of cooling
Rate of cooling $\left(R\right)\propto$ Difference in temperature of the body and suroundings i.e $(\theta -{\theta }_0)$
Rate of cooling of body $B_1$ is
$R_1=\frac{dQ}{dt}\propto ({\theta }_1-{\theta }_0)$ ...(i)
& Rate of cooling of body $B_2$ is
$R_2=\frac{dQ}{dt}\propto ({\theta }_2-{\theta }_0)$ ....(ii)

From equation (i) and (ii),
$\frac{R_1}{R_2}=\frac{{\theta }_1-{\theta }_0}{{\theta }_2-{\theta }_0}$
$\Rightarrow \frac{R_1}{R_2}=\frac{100-40}{80-40}\left\{\text{As per question}\right\}$
$\Rightarrow \frac{R_1}{R_2}=\frac{60}{40}=\frac{3}{2}$
Hence (a) is the correct answer.